Question

Fill in the blank to make the expression a perfect square.
y^2-6y+?

Answer 1

+9

Answer 2

:D let me know if u have any questions

Answer 3

thanks! :)

Answer 4

wait, how did yu get that

Answer 5

okay let me explain

Answer 6

\[y ^{2}-6y+x ^{2} = (y-x)^{2}\]

Answer 7

simplifying this equation will give you
x=3

Answer 8

so the missing value is 9

Answer 9

oh okay, I understand now :)

Answer 10

no problemo

Answer 11

if you have something like \[y^2-6y\] and you want to make it a perfect square, just take half of the coefficient of the \(y\) term, square it, and add that. Then you can rewrite it as \((y+a)^2\) where \(a =\) the value you squared.

\[y^2-6y + (\frac{-6}{2})^2 = y^2-6y + 9 = (y-3)^2\]

This is called "completing the square" for what should be fairly apparent reasons :-)

Answer 12

If you were solving an equation by this method, you would add the squared quantity to both sides of the equation to preserve the equality.

\[x^2+4x =0\]\[x^2+4x+(\frac{4}{2})^2 = 0+(\frac{4}{2})^2\]\[(x+2)^2 = 4\]\[\sqrt{(x+2)^2} =\pm \sqrt{4}\]\[x+2 = \pm2\]\[x=0,x=-4\]

Answer 13

as a check, solve by factoring:
\[x^2+4x=0\]\[x(x+4) = 0\]\[x=0\]\[x+4=0\]\[x=-4\]